ĐKXĐ: \(x^2\ge3\)
\(\Leftrightarrow x^2-3>\sqrt{x^2-3}\)
\(\Leftrightarrow\left(x^2-3\right)^2>x^2-3\)
\(\Leftrightarrow x^2-3>1\)
\(\Leftrightarrow x^2>4\Rightarrow\left[{}\begin{matrix}x< -2\\x>2\end{matrix}\right.\)
ĐKXĐ: \(x^2\ge3\)
\(\Leftrightarrow x^2-3>\sqrt{x^2-3}\)
\(\Leftrightarrow\left(x^2-3\right)^2>x^2-3\)
\(\Leftrightarrow x^2-3>1\)
\(\Leftrightarrow x^2>4\Rightarrow\left[{}\begin{matrix}x< -2\\x>2\end{matrix}\right.\)
A = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-3.\left(\dfrac{\sqrt{x}+3}{x-9}\right)\right):\left(\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-1\right)\)
a) Rut gon A
b) Tim GTNN cua A
Câu 1
a) A =3
B =\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\) (điêu kiện:x≥0;x≠4)
b)Tim giá trị của x để B=\(\dfrac{2}{3}\)A
tim x \(2\sqrt{x+3}=9x^2-x-4\)
\(P=\left(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\right).\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
a, rut gon
b, tim x de P=\(\sqrt{x}\)
B = \(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right).\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
Ket qua rut gon dk \(\sqrt{x}-1\) . Tim x de B<\(2\sqrt{x}+3\)
Cho biểu thức: P = (sqrt(x))/(sqrt(x) + 3) + (3sqrt(x))/(x - 9) a) Rút gọn biểu thức P. với x>=0;x ne9 . b) Tim giá trị của x để P = 2 ,
1. Tim x de \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}>\dfrac{1}{6}\) voi \(x>0\), \(x\ne1\), \(x\ne2\)
2. So sanh \(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\) voi \(\dfrac{1}{3}\)
tim x :
\(2\sqrt{36x+36}-\dfrac{1}{3}\sqrt{9x-9}-4\sqrt{4x-4}+\sqrt{x-1}=16\)
cho bieu thuc P= (\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-3}\) ): \(\frac{1}{x-1}\)
a) Tim dieu kien de P co nghia, rut gon bieu thuc P.
b) Tim cac so tu nhien x de \(\frac{1}{P}\)la so tu nhien
c) Tinh gia tri cua P voi x= 4-\(2\sqrt{3}\)
Giup mk vs mk dang can gap